In the given equation,potassium dichromate ac | vedclass

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(D) The equivalent weight of an oxidizing agent is calculated by the formula: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-facto

Vedclass · Accepted Answer

$M/6$

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(D) The equivalent weight of an oxidizing agent is calculated by the formula: $	ext{Equivalent weight} = \frac{	ext{Molecular weight}}{	ext{n-factor}}$.In the given redox reaction,$Cr_2O_7^{2-} + 14H^+ + 6e^- ightarrow 2Cr^{3+} + 7H_2O$,the total number of electrons gained by one mole of $K_2Cr_2O_7$ is $6$.Therefore,the n-factor for $K_2Cr_2O_7$ is $6$.Thus,the equivalent weight $= \frac{M}{6}$.

In the given equation,potassium dichromate acts as an oxidant in an acidic medium: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$. The equivalent weight of $K_2Cr_2O_7$ is (Molecular weight $= M$):

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